Math, asked by priyo1027, 1 year ago

what is the minimum value of 9tan^A+4cot^A
its urgent pleasee

Answers

Answered by shadowsabers03
2

We're actually given,

\begin{aligned}&9\tan^2A+4\cot^2A\quad\implies\quad 9\tan^2A+\dfrac{4}{\tan^2A}\end{aligned}

At the minimum value of this, the derivative of this will be at zero. So let's find what the derivative of it is.

We have to differentiate it with respect to  \tan A.

To avoid some confusion, let,

\leadsto\ x=\tan A\\\\\leadsto\ y=9\tan^2A+\dfrac{4}{\tan^2A}\ =\ 9x^2+\dfrac{4}{x^2}

So,

\dfrac{dy}{dx}\ =\ \dfrac{d}{dx}\left(9x^2+\dfrac{4}{x^2}\right)\ =\ \dfrac{d}{dx}\left(9x^2+4x^{-2}\right)\ =\ 18x-8x^{-3}

That is,

y'=18\tan A-\dfrac{8}{\tan^3A}

As we said earlier, to get  \min(y),\quad\quad\min(y')=0.

\implies\ \ 18\tan A-\dfrac{8}{\tan^3A}=0\\\\\\\implies\ \ 18\tan A=\dfrac{8}{\tan^3A}\\\\\\\implies\ \ \tan^4A=\dfrac{4}{9}\\\\\\\implies\ \ &\tan^2A=\dfrac{2}{3}

So, y becomes,

9\tan^2A+\dfrac{4}{\tan^2A}\ =\ 9\cdot\dfrac{2}{3}+\dfrac{4}{\left(\frac{2}{3}\right)}\ =\ \mathbf{12}

Hence 12 is the answer.

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